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xeze [42]
3 years ago
10

Which statement below is not a premise of the Kinetic-molecular theory of gases?

Chemistry
1 answer:
zhuklara [117]3 years ago
4 0

Answer:

d = all are premise of KMT.

Explanation:

A

Gas particles are extremely small and have relatively large distance between them.

B

Gas particles are continuously moving in random, straight-line motion as they are collide with each other and the container walls.

C

The average kinetic energy of gas particles is proportional to the temperature of gas.

All these are the premise of kinetic molecular theory. According to kinetic molecular theory, the particles pf gases are very small and randomly move in the available space. They move freely in straight line and colloid with each other and also the wall of container. This collision is elastic. As molecules strike with each other and walls of container their energy increases, and they gets warmer. So their kinetic energy is proportional to the temperature because mass of particle is constant.

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A star is observed from two positions of Earth in its orbit, in summer and winter. Which of these is the best method to calculat
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6 0
3 years ago
Dissolution of KOH, ΔHsoln:
swat32

Using Hess's law we found:

1) By <em>adding </em>reaction 10.2 with the <em>reverse </em>of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)   ΔH  (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the <em>total </em>change in enthalpy (ΔH).    

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq)   ΔHsoln    (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq)     ΔHneut     (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from <u>differents changes</u> in various <em>reactions </em>can be calculated as the <u>sum</u> of all the <em>enthalpies</em> of all those <em>reactions</em>.      

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq)    (10.3)

We need to <em>add </em>reaction 10.2 to the <u>reverse</u> of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

<u>Canceling</u> the KOH(s) from both sides, we get <em>reaction 10.3</em>:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)    (10.3)

2) The change in enthalpy for <em>reaction 10.3</em> can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

\Delta H = \Delta H_{soln} + \Delta H_{neut}

The enthalpy of <em>reaction 10.1 </em>(ΔHsoln) changed its sign when we reversed reaction 10.1, so:

\Delta H = \Delta H_{neut} - \Delta H_{soln}

Therefore, the ΔHsoln must be <u>subtracted</u> from ΔHneut to get the total change in enthalpy ΔH.

Learn more here:

  • brainly.com/question/2082986?referrer=searchResults
  • brainly.com/question/1657608?referrer=searchResults  

I hope it helps you!

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2 years ago
Please help me I would really appreciate it
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The number of students is your independent variable
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Pls help I’ll mark brainliest pls pls pls
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The tea was no longer hot or (brewed) so the 5th didn’t dissolve like the others because the tea was hot or warm enough anymore it cooled down. So the sugar won’t dissolve no more.
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